I'm new to digital radios and signal processing, so I apologize if this question is trivial but I haven't been able to find an answer here or by googling. Also, some terminology might be off, please feel free to refer me to correct sources or to correct my basic understanding.
Reading various sources (e.g. here), it seems to me that the I and Q components of a sample correspond to the complex representation of a portion of a sine wave described by $I \cdot \cos(2 \pi f t) + Q \cdot \sin(2 \pi f t)$ w.r.t. $t$, where $f$ denotes the frequency of interest. My question is, how does the receiver actually compute $I$ and $Q$ when a sample is needed?
Suppose that a sample is taken at a time $t$, I don't think that the receiver could just multiply the instantaneous strength $V$ (voltage?) of the incoming signal by $\cos(2\pi ft)$ and by $\sin(2 \pi f t)$ to recover $I$ and $Q$ (as the diagram in section "Receiver Side" of the linked article appears to suggests) since this would carry no more information than reporting $V$ itself.
Moreover, in principle, the incoming voltage from the antenna on the receiver side could be any continuous (and differentiable?) function $V(t)$... so how are $I$ and $Q$ recovered? Are they actually the values that minimize some error function between the incoming voltage and the function described by $I \cdot \sin(f) + Q \cdot \cos(f)$ over a length of time corresponding to some sampling interval $[t, t']$? E.g. something along the lines of: $$ I,Q = \arg\min_{I,Q \in \mathbb{R}}\int_{\tau=t}^{t'} \big( I \cdot \cos(2 \pi f \tau) + Q \cdot \sin(2 \pi f \tau) - V(\tau) \big)^2 \;\mbox{d}\tau \;\mbox{ ?} $$
Thank you!